Problem: What is the extraneous solution to these equations? $\dfrac{x^2}{x + 6} = \dfrac{x + 42}{x + 6}$
Answer: Multiply both sides by $x + 6$ $ \dfrac{x^2}{x + 6} (x + 6) = \dfrac{x + 42}{x + 6} (x + 6)$ $ x^2 = x + 42$ Subtract $x + 42$ from both sides: $ x^2 - (x + 42) = x + 42 - (x + 42)$ $ x^2 - x - 42 = 0$ Factor the expression: $ (x - 7)(x + 6) = 0$ Therefore $x = 7$ or $x = -6$ At $x = -6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -6$, it is an extraneous solution.